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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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Find all possible values of $p$ and $q$ for which $\cos^{-1}\sqrt p+\cos^{-1}\sqrt{1-p}=\large\frac{3\pi}{4}$

$\begin{array}{1 1}(a)\;p=0,q=\large\frac{1}{2}&(b)\;p \leq 2,q=1\\(c)\;0 \leq p\leq 1,q=\large\frac{1}{2}&(d)\;None\;of\;these\end{array}$

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1 Answer

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Step 1:
Given :
$\cos^{-1}\sqrt p+\cos^{-1}\sqrt{1-p}+\cos^{-1}\sqrt{1-q}=\large\frac{3\pi}{4}$
$\Rightarrow \cos^{-1}\sqrt p+\sin^{-1}\sqrt p+\cos^{-1}\sqrt{1-q}=\large\frac{3\pi}{4}$
Let $\cos^{-1}\sqrt{1-p}=\theta$
$\cos\theta=\sqrt{1-p}$
$\sin\theta=\sqrt p$
$\sin^{-1}x+\cos^{-1}x=\large\frac{\pi}{2}$
$\large\frac{\pi}{2}$$+\cos^{-1}\sqrt{1-q}=\large\frac{3\pi}{4}$
$\cos^{-1}\sqrt{1-q}=\large\frac{\pi}{4}$
$\sqrt{1-q}=\cos\large\frac{\pi}{4}$
$\Rightarrow \large\frac{1}{\sqrt 2}$
$1-q=\large\frac{1}{2}$
$q=\large\frac{1}{2}$
Step 2:
$\cos^{-1}\sqrt p\Rightarrow 0\leq \sqrt p\leq 1$
$\Rightarrow 0\leq p\leq 1$
$\cos^{-1}\sqrt{1-p}=0\leq \sqrt{1-p}\leq 1$
$\Rightarrow 0\leq p\leq 1$
$\therefore 0\leq p\leq 1$ and $q=\large\frac{1}{2}$
Hence (c) is the correct option.
answered Oct 22, 2013 by sreemathi.v
 

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